Cubic EquationsSeries from Integration by Parts
Introduction
The Taylor Series and Maclaurin Series are famous series that allow us to calculate values of transcendental functions. They are obtained from repeatedly taking the derivative of a fuction. So, I thought about creating series by repeatedly integrating a function. The resulting series may not be nice, however, it does allow us to come up with a series that can be used to approximate the antiderivative of functions that do not have an antiderivative.
Functions such as \( e^{x^{2}} \) do not have an antiderivative. Here, I present a method of deriving a series that can be used to get the ‘exact’ value of the integration limits of such functions by using integration by parts. By ‘exact’, I mean it is as accurate as the number of terms are used in the sum. Hence, it’s not just an approximation.
The method of integration by parts is well known: \( \int u \, dv = uv - \int v \, du \). If we iterate this, we get:
(1) \( \int u(x)v(x) \, dx = u(x)v_{1}(x) - u'(x)v_{2}(x) + u''(x)v_{3}(x) - u'''(x)v_{4}(x) \, + ... \)
The notation \( v_{n}(x) \) represents the nth antiderivative. If we let \( v(x) = 1 \) and \( u(x) = f(x) \), we have the following:
(2) \( \int f(x) \, dx = x\cdot f(x) - \frac{x^{2}}{2!}f'(x) + \frac{x^{3}}{3!}f''(x) \, - \) \( \frac{x^{4}}{4!}f'''(x) + \frac{x^{5}}{5!}f^{(4)}(x) \, - \) \( \frac{x^{6}}{6!}f^{(5)}(x) \, + ... \)
Example
Let’s try this on the function \( f(x) = e^{x^{2}} \), which does not have any algebraic or transcendental antiderivative. We will calculate the derivatives first:
(0) \( f(x) = e^{x^{2}} \)
(1) \( f'(x) = 2xe^{x^{2}} \)
(2) \( f''(x) = 2e^{x^{2}} + 4x^{2}e^{x^{2}} \)
(3) \( f'''(x) = 12xe^{x^{2}} + 8x^{3}e^{x^{2}} \)
(4) \( f^{4}(x) = 12e^{x^{2}} + 48x^{2}e^{x^{2}} + 16x^{4}e^{x^{2}} \)
(5) \( f^{5}(x) = 120xe^{x^{2}} + 160x^{3}e^{x^{2}} + 32x^{5}e^{x^{2}} \)
(6) \( f^{6}(x) = 120e^{x^{2}} + 720x^{2}e^{x^{2}} + 480x^{4}e^{x^{2}} + 64x^{6}e^{x^{2}} \)
Combining this with the equation in (2), we get:
(i) \( \int e^{x^{2}} \, dx = e^{x^{2}}\left( x - \frac{2}{3}x^{3} + \frac{4}{15}x^{5} - \frac{8}{105}x^{7} + ... + \frac{2^{n}}{1\cdot 3 \cdot 5 \cdot ... \cdot (2n+1)}x^{2n+1} \right) \)
Using the same process, we get the following for the negative in the exponent:
(ii) \( \int e^{-x^{2}} \, dx = e^{-x^{2}}\left( x + \frac{2}{3}x^{3} + \frac{4}{15}x^{5} + \frac{8}{105}x^{7} + ... + \frac{2^{n}}{1\cdot 3 \cdot 5 \cdot ... \cdot (2n+1)}x^{2n+1} \right) \)
Since the functions in (i) and (ii) do not have any integrals, these series are a good way to approximate the exact values when taking the limits.
The u Substitution
When it becomes difficult to find the nth derivatives, a change of variable can be helpful. For example, when integrating \( \int e^{n^{n}} \, dx \), we can let \( u = x^{n} \) so that \( dx = \frac{1}{n}u^{1/n-1} \, du \). Therefore, after the u-substitution, we have \( \int e^{u}\cdot \frac{1}{n}u^{1/n-1} \, du \). The n derivatives of \( u(x) \) are all \( e^{u} \), which can be factored out. We essentially end up with:
(i) \( \int e^{u} \cdot \frac{1}{n}u^{1/n-1} \, du = e^{u}\left( u^{1/n} - \frac{n}{n+1}u^{1/n+1} + \frac{n}{n+1}\cdot \frac{n}{2n+1}u^{1/n+2} - \frac{n}{n+1}\cdot \frac{n}{2n+1}\cdot \frac{n}{3n+1}u^{1/n+3} + ... \right) \)
Back substitution yields:
(ii) \( \int e^{x^{n}} \, dx = e^{x^{n}}\left( x - \frac{nx^{n+1}}{n+1} + \frac{n^2x^{2n+1}}{(n+1)(2n+1)} - \frac{n^3x^{3n+1}}{(n+1)(2n+1)(3n+1)} + ... \right) \, = \) \( xe^{x^{n}} \sum\limits_{i=1}^{\infty} \dfrac{(-n)^{i}x^{ni}}{\prod\limits_{j=0}^{i}(nj+1)} \)
Example
Does it work? Let’s try it with the definite integral \( \int_{0}^{2} e^{x^3} \, dx \). Wolfram Alpha gives the following result: \( \int_{0}^{2} e^{x^3} \, dx \approx 276.853... \)
The lower limit is just 0, so summing the series with the upper limit gives us:
\( \int_{0}^{2} e^{x^{3}} \, dx = \) \( 2e^{2^{3}} \sum\limits_{i=1}^{\infty} \dfrac{(-3)^{i}2^{3i}}{\prod\limits_{j=0}^{i}(3j+1)} \approx 276.853 \)
See summation result.Confirmed Series
(1) \( \int e^{x}\ln x \, dx = (e^{x}-1)\ln x - \left( x + \frac{x^{2}}{2\cdot 2!} + \frac{x^{3}}{3\cdot 3!} + \frac{x^{4}}{4\cdot 4!} + \frac{x^{5}}{5\cdot 5!} + ... \right) \, = \) \( (e^{x}-1)\ln x - \sum\limits_{i=1}^{\infty} \frac{x^{i}}{i\cdot i!} \)
(2) \( \int e^{-x}\ln x \, dx = (-e^{x}+1)\ln x + \left( x - \frac{x^{2}}{2\cdot 2!} + \frac{x^{3}}{3\cdot 3!} - \frac{x^{4}}{4\cdot 4!} + \frac{x^{5}}{5\cdot 5!} - ... \right) \, = \) \( (-e^{x}+1)\ln x + \sum\limits_{i=1}^{\infty} \frac{(-1)^{i}x^{i}}{i\cdot i!} \)
(3)\( \int \ln(\ln x) \, dx = (x-1)\ln(\ln x) - \left( \ln x + \frac{(\ln x)^2}{2\cdot 2!} + \frac{(\ln x)^3}{3\cdot 3!} + \frac{(\ln x)^4}{4\cdot 4!} + \frac{(\ln x)^5}{5\cdot 5!} + ... \right) \)